# Tag Info

3

While a single LHC particle wouldn't be doing much harm, being hit by the LHC beam would be certainly deadly and it would damage the machine badly. Any dense matter that comes into the LHC beam will instantly act as a beam dump. We have a very good idea about what happens in the LHC beam dump, see e.g. ...

7

Nothing happens obviously, when one high energy particle penetrates flesh as cosmic rays continuously impinge on us and some have the energies of the LHC. The cosmic rays reaching us are mainly muons and the damage they do is with electromagnetic scatters/ionisations in their path. The mean energy of muons reaching sea level is about 4 GeV. Muons, being ...

17

Amazingly this actually happened to a Russian scientist called Anatoli Bugorski. The beam basically just killed all the tissue it passed through. The symptoms were the relatively mundane ones expected from tissue death. The LHC has a much, much greater energy than the one that struck Bugorski, so it would cause a lot more heating and presumably burning of ...

8

Charged particle will make charge separation along it's path (ionization). This will cause malicious (wrong) chemical reactions occur in body, including DNA damage. The effect of these chemical reactions depend on their amount. Low amount can be self cured, while high amount will cause radiation sickness and probably death. This can be calculated, but also ...

2

What other variables I should know before it can be calculated? : I only know the values before the collision, for example the initial velocities VA,VB, initial temperatures, surface areas, etc. You need to know the $C_R$ coefficient of restitution, CR is the coefficient of restitution if it is 1 we have an elastic collision, if it is 0 we have a ...

0

Fundamentally, it comes down to the conservation of energy, since we already know the initial velocity of both objects, by simply measuring the velocities of the objects just after the collision (say ${v_1}$ for object 1 and $v_2$ for object 2), we can calculate the amount of energy converted to heat by simply using, $${ m_1v^2 = m_1v_1^2 + m_2v_2^2 + Heat} ... 0 Kinetic energy of the billiard ball will be maximum when the ball will hit the billiard ball in straight line with a given velocity.$$mv=mu+MU...........1$$And,$$mv^2=mu^2+MU^2............2$$Now the kinetic energy \frac{1}{2}MU^2 is a fraction of the kinetic energy \frac{1}{2}mv^2$$$\frac{1}{2}MU^2=k\frac{1}{2}mv^2............3$$From 2 and 3 find k in ... 0 The way I would approach this problem is through the impact parameter, b. You can find the definition on Wikipedia. The first thing to note is that for b > r + R, there is no collision. Here r and R are the radii of the two balls. For b less than this limit, you can use geometry to determine the direction of \mathbf{U}. This is because the ... 1 A "collision course" is a very fuzzy concept: if you are "barely going to hit" you are on a collision course but don't need a lot of deflection. However, let's assume for a moment a stationary earth, a meteorite of mass m at distance D, heading for earth of radius R with velocity v. The equations you need are conservation of angular momentum and ... 0 In an elastic collision the masses of both objects, the total kinetic energy, and the total linear momentum are conserved. The kinetic energy has contributions from the motions of the objects as well as their rotations. If we assume that no exchange between these two forms of kinetic energy occurs, i.e. that both forms are separately conserved, we have$$ ... 0 They will stick together if the collision is inellastic, which means you dissipate the maximum kinetic energy possible, while conserving momentum. As you might expect, metals, rubber, plastics, etc. all provide a partially elastic collision, since they are rigid. If the collision speed is high enough, the materials could always fuse (given that the objects ... 2 The equation,$v_A-v_B = \frac{4}{5}(u-2u/3)$is incorrect. The proper equation for the coefficient of restitution is given by,$v_A-v_B$=$e(u_B-u_A)$, where$u$and$v$are velocities along the line of impact. I believe you came across the somewhat incomplete statement that the relative speed of separation after the collision is$e$times the initial ... 1 Here is a general figure of an hard spheres collision drawn in the center of mass of the mass$m_2$before the collision. The black dot is attached to this frame. To solve the problem, you need to observe Conservation of energy:$m_1v_1^2=m_1(v'_1)^2+m_2(v'_2)^2$. Conservation of momentum:$m_1\vec v_1=m_1\vec v'_1+m_2\vec v'_2$Conservation of torque ... -2 The question that has illogical parameters that are probably not correct. It is unclear to what force exactly they are referring here. Consider the following question: A mosquito and a truck collide, both have the same velocity, they collide and and it takes one second for them to stop. That clearly illustrates the foolishness of the question. It sets ... 3 ...truck will experience larger force. But Newton's Third Law of motion says that 'to every action there is an equal and opposite reaction'. So the force experienced by the truck (M) should be same to that experienced by the car (m), but negative, isn't it? Third law states that momentum must be equal and opposite if both vehicles must come to a ... 0 The author of the question presumably expects you to notice these facts both have the same velocity they collide and stop The collision last for one second and to work out the total force acting on each vehicle from the change in their momentum. The answer given is arguable correct if the author is talking about the total force (i.e. not just the ... 1 They both experience the same force because of the impact, due to the Newton's third law, like you say. I think the question is not clear enough. If you assume there is no friction between the trucks and the ground, then you can use momentum considerations. I know this shouldn't be an answer, but I'm new and I can't post a comment, yet. 2 The main factor you should consider is impulse: if the force acts for a longer interval of time, it causes greater fractures in your glass, if it acts for a shorter interval of time, it causes less fractures. Again, it depends on the material of the glass and the projectile you are using. 2 The glass of the windows you see in movies are made of sugar or acrylic, that's why you can fly through the window (with a punch) without a scrape or get a 'bottle' of whisky smashed on your head without a headacke. That's why , especially with plastic you can see a clean hole. ...the stunt glass you see people jump through all the time in movies is not ... 1 I think I've get a better idea of what you are looking for now, thanks. As background for others in the future: classic ion-solid interaction theory dates back to the 1960s and is commonly called LSS theory after Linhard, Scharff, and Schiott who first formulated the concepts. It splits the energy loss mechanisms of the ion into two components, electronic ... 0 Since you are saying an object i am considering it to be a rigid particle.Now,since the particle strikes the surface as in your figure.it gives a downward force on the surface and hence the reaction is obviously upward. As you must be knowing normal reaction is perpendicular to the surface.So break the black arrow in componentsi.e you can imagine the ball to ... 1 I would proceed as follows: Assuming that the stored energy in the spring, compressed/extended a distance$d$from equilibrium, is given by$$E_{stored}=kd^2$$ Find stored energy, kinetic energy, and thus total energy at the moment of release; Find stored energy, and thus kinetic energy, just before the collision Step aside for a moment, and use ... 1 Remember that an inelastic collision means energy is lost but not momentum. Momentum is always conserved in collisions. So you have to figure out what kind of velocity the two joint masses have and use your knowledge of springs to compute the halt at the given position, velocity and mass. 0 So after a bit of research, i came up with this answer. As i had said before we know that in orbit, the mass of the satellites does not matter to find the velocity. so if we use: $$v = \sqrt{GM/R}$$ as M being the mass of earth i can find the final velocity of both satellites after collation. $$Vf = (m2v2 - m1v1)/ (m1 + m2)$$ I came up with 4407.87 m/s If ... 1 Emmy Noether discovered a fundamental connection between symmetries and conservation laws, embodied in her famous theorem. In simple terms, Noether's theorem is that For every symmetry in a physical system, there must be a conserved quantity. The proof requires neither Lorentz invariance nor causality. By applying Noether's theorem, we find that ... 1 We have to make a few assumptions here, because your question as posed is a bit incomplete. 1) collision is inelastic 2) satellites don't disintegrate on collision (see user58220's interesting observation) 3) there is no atmosphere (so drag is not an issue - let's see if the satellites end up in an orbit above the planet's surface) 4) you said "crash into ... 1 ... if the momentum of the system is conserved, the velocity of the centre of mass of the system should remain the same. True. ... the velocity of the centre of mass of the system has to be different after the collision for the kinetic energy to be different. False. So, how can there be a change in kinetic energy of the ... 1 if mass is assumed to be constant, the velocity of the centre of mass of the system has to be different after the collision for the kinetic energy to be different. However, if the momentum of the system is conserved, the velocity of the centre of mass of the system should remain the same. 1) mass is not constant and velocity is different: in ... 0 a. Use kinematic equation for constant acceleration: v^2=u^2+2as Here v, the final velocity is 0 because the car comes to a stop, u is 60 mph, s is 1.5 m. Therefore a= u^2/2s. b. Just substitute u=90 mph and evaluate a in the same way. Also to get the answers in g, just divide a by 9.8 m/s^2 Note: a will be negative 1 Introducing a side issue to this question: Using$v$for the orbital velocity at$1000$km altitude, the total kinetic energy of the two satellites just before the collision is $$KE_{\text{Before}}=\frac12400v^2+\frac12100v^2=250v^2$$and using the final velocity of John Rennie above, the total kinetic energy of the combined satellites just after the ... 8 A simple counterexample: Imagine two particles with opposite direction and equal speed. The center of mass does not move, yet the kinetic energy of the system is non-zero. Now let both particles come to rest (by friction, hitting a wall, whatever). The kinetic energy is now zero, and total momentum has been conserved, while energy is not. The crucial ... 5 Unless I'm missing an easy way to do this problem it seems a surprisingly hard one. This diagram shows the problem (I've exaggerated the altitude of the satellite to make the diagram clearer): The satellites are in circular orbits (dotted line) at a distance$r$from the centre of the Earth, so their orbital velocity as (as you say):$\$ v = ...

0

I can only assume that the satellites are assumed to meet head-on (rather than at an angle), and that the collision is modeled as inelastic (rather than an explosive collision where pieces fly everywhere). If so, you have the velocity of each and the mass of each. You can use the conservation of momentum to find the new velocity of the combined-mass object ...

Top 50 recent answers are included